ch02-solutions.pdf

# ch02-solutions.pdf - PHYS2405 Chapter 2 Standing Waves in...

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Unformatted text preview: PHYS2405 Chapter 2: Standing Waves in Ropes Solutions Spring 2019 At t = 0 s DISPLACEMENT (ft) 3 2 1 0 −1 −2 At x = 3 ft DISPLACEMENT (ft) −3 0 2 4 x (feet) 6 8 10 2 1 0 −1 −2 −3 0 5 t (seconds) 10 15 Figure 1 A standing wave of a string tied down at = 0 and = 10 f t. The top graph shows a snapshot of the string (height versus position) at time = 0. The lower graph (height versus time) shows how the point at = 3 f t behaves with time. Problems 1-11 refer to this figure. Question 1 a. 3 ft. What is the wavelength of the standing wave in Figure 1? b. 4 ft. c. 5 ft. d. 10 ft. Solution Since the top graph in Figure 1 shows displacement versus distance, we can directly read the wavelength from it. One full cycle of the wave takes us from = 0 to = 4 f t. So this is the wavelength: = 4 f t. Question 2 a. 3 ft. What is the amplitude of the standing wave in Figure 1? b. 4 ft. c. 5 ft. d. 10 ft. e. none of the above. Solution The amplitude of a wave is the ‘height’ of the wave: from zero displacement to maximum displacement. From either graph in Figure 1, the maximum displacement is 3 f t. Question 3 a. 4 s. What is the period of the standing wave in Figure 1? b. 6 s. c. 8 s. d. 10 s. e. none of the above. Solution The bottom graph in Figure 1 shows displacement versus time, we can read the wave’s period by determining the time for one complete cycle. From the graph, at = 0 we have a trough and the next trough occurs at = 8 s. So the period is = 8 s. Stephens PHYS2405 Question 4 Chapter 2 Solutions, page 2 of 4 Spring 2019 What is the frequency of the standing wave in Figure 1? a. 1/16 Hz. b. 1/8 Hz. c. 8 Hz. d. 16 Hz. e. none of the above. Solution Now that we know the period, we can calculate the frequency: = Question 5 1 1 1 1 = = 8s 8 s ⇒ 1 = 8 Hz. What is the wave velocity of the standing wave in Figure 1? a. 1/2 ft/s. b. 2 ft/s. c. 8 ft/s. d. 16 ft/s. e. none of the above. Solution We now know the wave amplitude and frequency. So we can calculate the wave velocity from = = ( 18 1/s) (4 ft) Question 6 ⇒ = 1 ft 2 s . How many nodes are there in the standing wave in Figure 1? a. 0. b. 4. c. 6. d. 7. e. 10. Solution To find the nodes, we need to consider a snapshot of the string: what the shape of the string is at a given instant. So from the top graph, we can see that there are nodes at = 0, 2, 4, 6, 8, and 10 ft. So there is a total of 6 nodes. By the way, you do count the nodes at the ends of the string. Question 7 How many antinodes are there in the standing wave in Figure 1? a. 0. b. 2. c. 3. d. 5. e. 8. Solution Again referring to the top graph, the antinodes occur at = 1, 3, 5, 7, and 9 ft. So there is a total of 5 nodes. For a string fixed at each end, the number of antinodes is always one less than the number of nodes. Question 8 Which harmonic is the standing wave in Figure 1? a. none of the others. b. third harmonic. c. fourth harmonic. d. fifth harmonic. e. sixth harmonic. Solution For a string fixed at each end, the fundamental (first harmonic) has a single antinode at the middle of the string. This string has five antinodes. so it corresponds to the 5th-harmonic. Another method to determine the harmonic number is (for a string fixed at both ends) is to count the number of half-wavelengths. Since we have five (from 0-2 ft, 2-4 ft, 4-6 ft, 6-8 ft, and 8-10 ft), this is the fifth harmonic. If this string was free at one end, we would count the number of quarter-wavelengths to determine the harmonic number. Question 9 Which overtone is the standing wave in Figure 1? a. none of the below. b. third overtone. c. fourth overtone. Stephens d. fifth overtone. e. sixth overtone. PHYS2405 Chapter 2 Solutions, page 3 of 4 Spring 2019 Solution For a string fixed at each end, the overtone number is one less than the harmonic number. So this corresponds to the fourth overtone. If the string was free at one end, the relationship between harmonic number and overtone number is not as simple. Question 10 a. 1/40 Hz. What is the frequency of the second harmonic of the string in Figure 1? b. 1/20 Hz. c. 1/2 Hz. d. 80 Hz. e. none of the above. Solution There are several ways to do this. We know that the top graph in Figure 1 corresponds to the fifth harmonic and it 1 has a frequency of = 8 Hz. So in terms of the frequency of the fundamental 1 , we have 1/8 Hz 5 = 51 = 18 Hz → 1 = 5 = ⇒ 1 = 401 Hz. 5 5 So the second harmonic must have a frequency 2 = 21 = 2 ( 401 Hz) ⇒ 2 = 1 20 Hz. Another way is to realize that for the second harmonic, the wavelength equals the string’s length. This means we 1 have 2 = = 10 f t. We found the wave velocity above to be = 2 ft/s. So we can determine the frequency for this harmonic as 1/2 ft/s 1 = → = 2 2 → 12 ft/s = 2 (10 f t) → 2 = ⇒ 2 = 20 Hz. 10 f t Question 11 a. 3 ft. What wavelength corresponds to the frequency of Question 10? b. 4 ft. c. 5 ft. d. 10 ft. Solution Since we know that Question 10 refers to the second harmonic and that the second harmonic has a wavelength equal to the string’s length, we immediately know that = = 10 f t. Question 12 A piece of wire is cut into two pieces, A and B, which are then tightly stretched and mounted between two rigid walls. A and B have the same stretched lengths, but A is stretched more tightly than B. Which of the following quantities will always be larger for waves on A than for waves on B? a. amplitude of the wave. b. frequency of the first harmonic. c. wave velocity. d. wavelength of the first harmonic. e. both b and c. Solution Let’s first consider what is always the same. Since the strings have the same lengths, then their fundamentals must always have the same wavelength: = . Now increasing the tension of string A means that it will produce sounds of higher frequency (think of a guitar or violin, or stretching a rubber band between your fingers). So this means that > , regardless of which harmonic we consider (as long as we consider the same harmonic for each string, i.e., first vs. first, second vs. second, etc., and not first vs. second, etc.). We can write this as / > 1. This means that choice b must be correct. Now we need to do a little reasoning. We know that = . But let’s rewrite this as = / and use what we know: = Stephens → = → = >1 ⇒ > 1. PHYS2405 Chapter 2 Solutions, page 4 of 4 Spring 2019 So what did we just do? Since the wavelengths are equal, the ratio / must also be equal for each string. This means that the ratio of the frequencies, / , has the same value as the ratio of the wave velocities, / . Since we know that is greater than , this means that / > 1, or > . So choice c must also be correct. That leaves only one correct choice. Question 13 A piece of wire is cut into two pieces, A and B, stretched to the same tension, and mounted between two rigid walls. Segment A is longer than segment B. Which of the following quantities will always be larger for waves on A than for waves on B? Solution Since the string tensions are the same, we know that the wave velocities and frequencies are the same (see the previous question for justification). That is, = and = . The problem also states that > . This directly implies that > > , as long as we compare equal harmonics of the two strings. With these results, there is only one correct choice. a. amplitude of the wave. b. frequency of the first harmonic. c. wave velocity. d. wavelength of the first harmonic. Question 14 A 4-meter-long string fixed at both ends has a wave velocity of 40 m/s. Which of the following is not an allowed wavelength for a standing waves on the string? a. 1/2 m. b. 2. c. 8/3 m. d. 10 m. Solution We know the length of the string, = 4 m, and the wave velocity, = 40 m/s. For the fundamental (first harmonic), we know the wavelength 1 is twice the string length: 1 = 2 = 8 m. Because all harmonics are possible for a string fixed at both ends, = 1 for the frequencies translates into = 1 /. So possible wavelengths must be eight divided by any whole number, i.e., 8/1 = 8 m, 8/2 = 4 m, 8/3 = 8/3 m, etc. There is no way we can divide eight by a number greater than one and get 10. So, there is only one possible answer. Question 15 A 5-meter-long string fixed at one end only has a wave velocity of 20 m/s. Which of the following is not allowed frequency for standing waves on this string? a. 1 Hz. b. 2 Hz. c. 3 Hz. d. 9 Hz. e. 101 Hz. Solution Since the string is fixed at only one end, we can only have odd harmonics. For the fundamental, we know that 1 = 4 = 20 m. This implies a frequency 1 = /1 = 1 Hz. So possible frequencies for the harmonics are 1 = 1 Hz, 3 = 3 Hz, 5 = 5 Hz, etc — no even numbers. Stephens ...
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• Spring '19
• STEPHENS
• Wavelength, Standing wave, Stephens

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